Let $k$ be the ratio of the small cone’s height to the big cone’s height. For similar solids, how does the volume scale in terms of $k$?

The frustum’s volume is the original cone’s volume minus the small cone’s volume. Write this in terms of $k^3$ and set it equal to $\tfrac{7}{8}$ of the original volume.

Once you know $k$, remember that $k = \dfrac{h}{15}$. How can you use this to find $h$, the height of the small cone?

In Desmos, type the equation 1 - k^3 = 7/8. Then either use the wrench menu and the "Solve for" feature (if available) or graph y = 1 - k^3 and y = 7/8 and find the intersection point’s k-coordinate.

Once you have the value of $k$ from Desmos, enter a new expression 15 * k. The output of this expression is the height of the smaller cone in centimeters.

The plane cuts the cone parallel to its base, so the top piece is a smaller cone that is similar to the original cone.

Let:

• $H = 15$ be the height of the original cone.
• $h$ be the height of the smaller cone removed.

Because the cones are similar, the ratio of their heights equals the ratio of any other corresponding linear dimensions (like radii). So we can write:

for some scale factor $k$ between 0 and 1.

For similar 3D shapes, volumes scale by the cube of the linear scale factor.

Let $V$ be the volume of the original cone, and $V_s$ be the volume of the small cone.

Then:

The frustum is what’s left after removing the small cone, so its volume is:

We are told that the frustum’s volume is $\tfrac{7}{8}$ of the original cone’s volume:

From Step 2, we also have $V_{\text{frustum}} = (1 - k^3) V$.

Set these equal and solve for $k$:

Take the cube root of both sides:

Recall that $k = \dfrac{h}{H}$ and $H = 15$.

So:

Multiply both sides by 15:

Therefore, the height of the smaller cone that was removed is $7.5$ centimeters.